Question

1.The reaction of aluminum metal (Al) with oxygen (O2) forms Al2O3. Write a balanced equation for this redox reaction. Write two half reactions to show how many electrons are gained or lost by each species.

Redox Reaction:

Oxidation Reaction:

Reduction Reaction:

2. Be sure to answer all parts. Hydrocarbons are compounds that contain only C and H atoms. When a hydrocarbon reacts with O2, CO2 and H2O are formed. Write a balanced equation for the combustion of the following hydrocarbon, a high-octane components of gasoline. Do not include states of matter in your answer. C8H18 (isooctane)

3.One dose of an antacid contains 750. mg each of Mg(OH)2 and Al(OH)3. How many moles of each compound are contained in a single dose? Use proper scientific notaion to report the answers.

---- × 10 mol Mg(OH)2

----× 10 mol Al(OH)

Answer 1

Answer 1):-

Redox reaction : 4 Al + 3 O₂ 2 Al₂O₃

Oxidation Reaction : 4 Al 4 Al³⁺ + 12e^-

Reduction Reaction​​​​​​ : 3 O​​​​​​​₂  + 12e^- 6 O^2-

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Answer 2):-

The balanced reaction for the complete combustion of isooctane :-

2 C₈H₁₈ + 25 O​​​​​​₂ 16 CO₂ + 18 H₂O

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Answer 3):-

Mass of Mg(OH)₂ = 750 mg = 0.75 g

Mass of Al(OH)₃ = 750 mg = 0.75 g

First, we have to find the molar mass of both Mg(OH)₂ and Al(OH)₃ .

Molar mass of Mg(OH)₂ = ( 1 × Atomic mass of Mg + 2 × Atomic mass of O + 2 × Atomic mass of H) = ( 1 × 24.305 + 2 × 15.999 + 2 × 1.008) = 58.319 g/mol

​​​​​​Molar mass of Al(OH)₃ = (1 × Atomic mass of Al + 3 × Atomic mass of O + 3 × Atomic mass of H) = (1 × 26.981 + 3 × 15.999 + 3 × 1.008) = 78 g/mol

Now,

Number of moles of Mg(OH)₂ = Mass ÷ Molar mass= (0.75 g ÷ 58.319 g/mol)= 1.29 × 10^-2mol

Number of moles of Al(OH)₃ = Mass ÷ Molar mass = (0.75 g ÷ 78 g/mol) = 9.62 × 10^-3 mol

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